To get to the point - what's the defining differences between them? Alas, my current understanding of a spinor is limited. All I know is that they are used to describe fermions (?), but I'm not sure why?

Although I should probably grasp the above first, what is the difference between Dirac, Weyl and Majorana spinors? I know that there are similarities (as in overlaps) and that the Dirac spinor is a solution to the Dirac equation etc. But what's their mathematical differences, their purpose and their importance?

(It might be good to note that I'm coming from a string theory perspective. Plus I've exhausted Wikipedia here.)


2 Answers 2


Recall a Dirac spinor which obeys the Dirac Lagrangian

$$\mathcal{L} = \bar{\psi}(i\gamma^{\mu}\partial_\mu -m)\psi.$$

The Dirac spinor is a four-component spinor, but may be decomposed into a pair of two-component spinors, i.e. we propose

$$\psi = \left( \begin{array}{c} u_+\\ u_-\end{array}\right),$$

and the Dirac Lagrangian becomes,

$$\mathcal{L} = iu_{-}^{\dagger}\sigma^{\mu}\partial_{\mu}u_{-} + iu_{+}^{\dagger}\bar{\sigma}^{\mu}\partial_{\mu}u_{+} -m(u^{\dagger}_{+}u_{-} + u_{-}^{\dagger}u_{+})$$

where $\sigma^{\mu} = (\mathbb{1},\sigma^{i})$ and $\bar{\sigma}^{\mu} = (\mathbb{1},-\sigma^{i})$ where $\sigma^{i}$ are the Pauli matrices and $i=1,..,3.$ The two-component spinors $u_{+}$ and $u_{-}$ are called Weyl or chiral spinors. In the limit $m\to 0$, a fermion can be described by a single Weyl spinor, satisfying e.g.


Majorana fermions are similar to Weyl fermions; they also have two-components. But they must satisfy a reality condition and they must be invariant under charge conjugation. When you expand a Majorana fermion, the Fourier coefficients (or operators upon canonical quantization) are real. In other words, a Majorana fermion $\psi_{M}$ may be written in terms of Weyl spinors as,

$$\psi_M = \left( \begin{array}{c} u_+\\ -i \sigma^2u^\ast_+\end{array}\right).$$

Majorana spinors are used frequently in supersymmetric theories. In the Wess-Zumino model - the simplest SUSY model - a supermultiplet is constructed from a complex scalar, auxiliary pseudo-scalar field, and Majorana spinor precisely because it has two degrees of freedom unlike a Dirac spinor. The action of the theory is simply,

$$S \sim - \int d^4x \left( \frac{1}{2}\partial^\mu \phi^{\ast}\partial_\mu \phi + i \psi^{\dagger}\bar{\sigma}^\mu \partial_\mu \psi + |F|^2 \right)$$

where $F$ is the auxiliary field, whose equations of motion set $F=0$ but is necessary on grounds of consistency due to the degrees of freedom off-shell and on-shell.

  • $\begingroup$ So am I right in saying all Majorana spinors are Weyl spinors, but not the other way around? And Weyl spinors, and therefore Majorana spinors as well, are subsets of Dirac spinors? $\endgroup$
    – Phibert
    Mar 11, 2014 at 15:13
  • $\begingroup$ I think you got it the other way around. All Majorana spinors are constructed from Weyl spinors, but Weyl spinors are not Majorana spinors. $\endgroup$
    – user32361
    Mar 11, 2014 at 15:15
  • 3
    $\begingroup$ This is what I said? $\endgroup$
    – Phibert
    Mar 12, 2014 at 0:09

After you will learn more about spinors, you will see that all spinors belong to the $\left(\frac{1}{2}, 0\right) + \left( 0, \frac{1}{2}\right)$ representation of the $SL(2,C)$ group, which is the double cover of the lorentz group $SO(3,1)$. The idea is to find representations of a simply connected covering group which in this case is $SL(2,C)$, the local structure given by the lie algebraic commutation relation remains the same.

Spinorial equations allow to extract Lorentz-invariant subspaces in the overall space of $\left(\frac{1}{2}, 0\right) + \left( 0, \frac{1}{2}\right)$ representation.

Both Dirac and Majorana spinors belong to $\left(\frac{1}{2}, 0\right) + \left( 0, \frac{1}{2}\right)$ representation of $SL(2,C)$ group, but they are only subspaces of it. For instance, Majorana spinors are all electrically neutral (i.e. remain invariant under charge conjugation). Similarly, Dirac spinors are "magnetically neutral".

Weil spinors belong to either $\left(\frac{1}{2}, 0\right)$ or $\left( 0, \frac{1}{2}\right)$ subspaces. Unlike Dirac and Majorana spinors, they might be considered as 2-component spinors. But this is also a limitation, because some special Lorentz transformations cannot be applied to these spinors.

  • 1
    $\begingroup$ Your first line is another issue I've been struggling to comprehend for a while now. Care to explain what you mean by $(\frac{1}{2},0)+(0,\frac{1}{2})$ and $SL(2,C)$? Representations just seem beyond me at the moment. $\endgroup$
    – Phibert
    Mar 11, 2014 at 16:07
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    $\begingroup$ @user13223423: The $so(3,1)$ lie algebra decomposes as $su(2)_L \oplus su(2)_R$. So any representation of $su(3,1)$ must be a tensor product of representations of the two subalgebras. Weyl spinors are in the "fundamental" rep of one of the $su(2)$ while they're in the trivial representation of the other. That is denoted by $({\frac{1}{2}}_L , 0_R)$ or the other way around. The Dirac and Majorana representations happen to be a linear combination of the two Weyl representations (with maybe other conditions). $\endgroup$
    – Siva
    Mar 11, 2014 at 21:59
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    $\begingroup$ There is a good explanation about representations of $\mathfrak{so}(3,1)$ in Weinberg's book. @Siva, you mean direct sum, not tensor product. The direct sum of (1/2,0) and (0,1/2) is the Dirac spinors, but their tensor product is (1/2,1/2) which is the 4-vector representation. As for the special Lorentz transformations, parity takes left-handed Weyl spinors to right-handed and the other way around. $\endgroup$ Apr 4, 2014 at 9:58
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    $\begingroup$ Whoops! That was a bad slip on my part. Thanks for the correction @RobinEkman. It is indeed the direct sum. $\endgroup$
    – Siva
    Apr 4, 2014 at 14:16
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    $\begingroup$ @Murod: Shoudn't your first sentence read double cover of $SO(3,1)$, spinors are representations of $SL(2,C)$ which is the double cover of $SO(3,1)$ - the lorentz group also called projective representations of the lorentz group for the same reason. $\endgroup$
    – user7757
    Jul 11, 2014 at 12:50

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